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We are asked to compute the average velocity of a stone over various time intervals given that the stone's height \( h(t) \) is described by:

\[
h(t) = 33t - 4.9t^2
\]

The formula for average velocity over a time interval \([t_1, t_2]\) is:

\[
\bar{v} = \frac{h(t_2) - h(t_1)}{t_2 - t_1}
\]

Now, let’s compute the average velocity for each of the given time intervals: 
- \([1, 1.01]\)
- \([1, 1.001]\)
- \([1, 1.0001]\)
- \([0.9999, 1]\)
- \([0.999, 1]\)

I'll calculate each of these next.The average velocities for the given time intervals are:

1. For \([1, 1.01]\): \(23.151 \, \text{m/s}\)
2. For \([1, 1.001]\): \(23.195 \, \text{m/s}\)
3. For \([1, 1.0001]\): \(23.200 \, \text{m/s}\)
4. For \([0.9999, 1]\): \(23.200 \, \text{m/s}\)
5. For \([0.999, 1]\): \(23.205 \, \text{m/s}\)

Would you like a detailed breakdown of how these were calculated? Let me know if you have any questions!

Here are 5 related questions you might find interesting:
1. How does the instantaneous velocity differ from the average velocity?
2. What is the physical meaning of negative acceleration in this problem?
3. How can we derive the instantaneous velocity function from the height function?
4. What would be the stone's velocity at \( t = 1 \) exactly?
5. How would the time intervals affect the accuracy of estimating instantaneous velocity?

**Tip:** Average velocity over a small interval closely approximates the instantaneous velocity at the midpoint of the interval.

A stone is tossed into the air from ground level with an initial velocity of 33 m/s. Its height at time t is h(t) = 33t - 4.9t^2 m. Compute the stone's average velocity over the time intervals [1, 1.01], [1, 1.001], [1, 1.0001], and [0.9999, 1], [0.999, 1]. (Use decimal notation. Give your answers to three decimal places.)

This problem involves calculating the average velocity of a stone tossed into the air, given the height function h(t) = 33t - 4.9t^2. Detailed step-by-step solutions for time intervals [1, 1.01], [1, 1.001], [1, 1.0001], and [0.9999, 1], [0.999, 1].

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